LOADING . . .
જવાબ :
જવાબ : Nonterminating nonrepeating.
જવાબ : Algorithm
398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527
HCF of 391, 425, 527 = 17
જવાબ : 2 × 7^{2}
જવાબ : HCF of 408 and 1032 is 24.
1032 × 2 + 408 × (p) = 24
408p = 24 – 2064
p = 5
જવાબ : We know,
1st number × 2nd number = HCF × LCM
⇒ 27 × 2nd number = 9 × 459
⇒ 2nd number = = 153
જવાબ : 13 = 1 × 13; 17 = 1 × 17
HCF = 1 and LCM = 13 × 17 = 221
જવાબ : LCM (3 × 5^{2}, 3^{2} × 7^{2}) = 3^{2} × 5^{2} × 7^{2} = 9 × 25 × 49 = 11025
જવાબ : It is given that on dividing 70 by the required number, there is a remainder 5.
This means that 70 – 5 = 65 is exactly divisible by the required number.
Similarly, 125 – 8 = 117 is also exactly divisible by the required number.
65 = 5 × 13
117 = 3^{2} × 13
HCF = 13
Required number = 13
જવાબ : 96 = 2^{5} × 3
360 = 2^{3} × 3^{2} × 5
LCM = 2^{5} × 3^{2} × 5 = 32 × 9 × 5 = 1440
જવાબ : 865 > 255
865 = 255 × 3 + 100
255 = 100 × 2 + 55
100 = 55 × 1 + 45
55 = 45 × 1 + 10
45 = 10 × 4 + 5
10 = 5 × 2 + 0
The remainder is 0.
HCF = 5
જવાબ : Let x = …(i)
100x = 612. …(ii)
…[Multiplying both sides by 100]
Subtracting (i) from (ii),
99x = 606
x = =
Denominator = 33
Prime factorisation = 3 × 11
જવાબ : y = 5 × 13 = 65
x = 3 × 195 = 585
જવાબ : Let us assume, to the contrary, that 2 + 3√5 is rational.
So that we can find integers a and b (b ≠ 0).
Such that 2 + 3√5 = , where a and b are coprime.
Rearranging the above equation, we get
Since a and b are integers, we get is rational and so √5 is rational.
But this contradicts the fact that √5 is irrational.
So, we conclude that 2 + 3√5 is irrational.
જવાબ : Let us assume, to the contrary, that 3√7 is rational.
That is, we can find coprime a and b (b ≠ 0) such that 3√7 =
Rearranging, we get √7 =
Since 3, a and b are integers, is rational, and so √7 is rational.
But this contradicts the fact that √7 is irrational.
So, we conclude that 3√7 is irrational.
જવાબ : 17 × 5 × 11 × 3 × 2 + 2 × 11 …(i)
= 2 × 11 × (17 × 5 × 3 + 1)
= 2 × 11 × (255 + 1)
= 2 × 11 × 256
Number (i) is divisible by 2, 11 and 256, it has more than 2 prime factors.
Therefore (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.
જવાબ : 4^{n} = (2^{2})n = 2^{2n}
The only prime in the factorization of 4^{n} is 2.
There is no other prime in the factorization of 4^{n} = 2^{2n}
(By uniqueness of the Fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of 4^{n} for any n.
Therefore, 4^{n} does not end with the digit zero for any natural number n.
જવાબ : No, LCM = Product of the highest power of each factor involved in the numbers.
HCF = Product of the smallest power of each common factor.
We can conclude that LCM is always a multiple of HCF, i.e., LCM = k × HCF
We are given that,
LCM = 175 and HCF = 15
175 = k × 15
⇒ 11.67 = k
But in this case, LCM ≠ k × HCF
Therefore, two numbers cannot have LCM as 175 and HCF as 15.
જવાબ : Irrational No.
જવાબ : Neither Rational nor Irrational no.
જવાબ : Irrational No.
જવાબ : Rational No.
જવાબ : Irrational No.
જવાબ : Rational No.
જવાબ : Irrational No.
જવાબ : RationalNo.
જવાબ : Rational
જવાબ : quotient
જવાબ : Euclid
જવાબ : remainder
જવાબ : Composite
જવાબ : Prime No.
જવાબ : Terminating
જવાબ : (√2−√3)(√3+√2)(2−3)(3+2)
=(√2−√3)(√2+√3)=(2−3)(2+3)
=23=1
So, it is rational number
જવાબ : 29029=29×13×11×729029=29×13×11×7
580=29×5×4580=29×5×4
HCF=29
LCM=29×13×11×7×4×5=58058029×13×11×7×4×5=580580
જવાબ : 120=2×2×3×2×5120=2×2×3×2×5
144=2×2×3×2×2×3144=2×2×3×2×2×3
HCF=23×3=24HCF=23×3=24
LCM=720
જવાબ : 27=3×3×327=3×3×3
81=3×3×3×381=3×3×3×3
HCF=27
LCM=81
જવાબ : 2 & 5
જવાબ : axb
જવાબ : 54
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : True
જવાબ : True
જવાબ : False
જવાબ : Let us assume, to the contrary, that √5 is rational.
So, we can find integers p and q (q ≠ 0), such that
√5 = , where p and q are coprime.
Squaring both sides, we get
5 =
⇒ 5q^{2} = p^{2} …(i)
⇒ 5 divides p^{2}
5 divides p
So, let p = 5r
Putting the value of p in (i), we get
5q^{2} = (5r)^{2}
⇒ 5q^{2 }= 25r^{2}
⇒ q^{2} = 5r^{2}
⇒ 5 divides q^{2}
5 divides q
So, p and q have atleast 5 as a common factor.
But this contradicts the fact that p and q have no common factor.
So, our assumption is wrong, is irrational.
√5 is irrational, 3 is a rational number.
So, we conclude that 3 + √5 is irrational.
જવાબ : Let us assume to the contrary, that 3 + 2√3 is rational.
So that we can find integers a and b (b ≠ 0).
Such that 3 + 2√3 = , where a and b are coprime.
Rearranging the equations, we get
Since a and b are integers, we get is rational and so √3 is rational.
But this contradicts the fact that √3 is irrational.
So we conclude that 3 + 2√3 is irrational.
જવાબ : 9 = 3^{2}, 12 = 2^{2} × 3, 15 = 3 × 5
LCM = 2^{2} × 3^{2} × 5 = 4 × 9 × 5 = 180 minutes or 3 hours
They will next toll together after 3 hours.
જવાબ : To find the maximum capacity of a container which can measure the petrol of each tanker in the exact number of times, we find the HCF of 850 and 680.
850 = 2 × 5^{2} × 17
680 = 2^{3} × 5 × 17
HCF = 2 × 5 × 17 = 170
Maximum capacity of the container = 170 liters.
જવાબ : To find the length of the longest rod that can measure the dimensions of the room exactly, we have to find HCF.
L, Length = 8 m 50 cm = 850 cm = 2^{1} × 5^{2} × 17
B, Breadth = 6 m 25 cm = 625 cm = 5^{4}
H, Height = 4 m 75 cm = 475 cm = 5^{2} × 19
HCF of L, B and H is 5^{2} = 25 cm
Length of the longest rod = 25 cm
જવાબ : To find the time when the clocks will next ring together,
we have to find LCM of 4, 12 and 20 minutes.
4 = 2^{2}
12 = 2^{2} × 3
20 = 2^{2} × 5
LCM of 4, 12 and 20 = 2^{2} × 3 × 5 = 60 minutes.
So, the clocks will ring together again after 60 minutes or one hour.
જવાબ : Since the books are to be distributed equally among the students of Section A and Section B. therefore, the number of books must be a multiple of 48 as well as 60.
Hence, required num¬ber of books is the LCM of 48 and 60.
48 = 2^{4} × 3
60 = 2^{2} × 3 × 5
LCM = 2^{4} × 3 × 5 = 16 × 15 = 240
Hence, required number of books is 240.
જવાબ :
જવાબ : Given numbers are 650 and 1170.
1170 > 650
1170 = 650 × 1 + 520
650 = 520 × 1 + 130
520 = 130 × 4 + 0
HCF = 130
The required largest number is 130.
જવાબ : 867 is greater than 255. We apply the division lemma to 867 and 255, to get
867 = 255 × 3 + 102
We continue the process till the remainder is zero
255 = 102 × 2 + 51
102 = 51 × 2 + 0, the remainder is zero.
HCF = 51
જવાબ : Let us assume, to the contrary, that 3 + 2√5 is rational
So that we can find integers a and b (b ≠ 0), such that
3 + 2 √5 = , where a and b are coprime.
Rearranging this equation, we get
Since a and b are integers, we get that – is rational and so √5 is rational.
But this contradicts the fact that √5 is irrational.
So we conclude that 3 + 2√5 is irrational.
જવાબ : (a) Find the maximum number of parallel rows of each class for the seating arrange¬ment.
(b) Also, find the number of students of class IX and also of class X in a row.
(c) What is the objective of the school administration behind such an arrangement?
Solution:
104 = 2^{3} × 13
96 = 2^{5} × 3
HCF = 2^{3} = 8
(a) Number of rows of students of class X = = 13
Number maximum of rows class IX = = 12
Total number of rows = 13 + 12 = 25
(b) No. of students of class IX in a row = 8
No. of students of class X in a row = 8
(c) The objective of school administration behind such an arrangement is fair and clean examination, so that no student can take help from any other student of his/her class.
જવાબ : 1st vessel = 720 ml; 2nd vessel = 405 ml
We find the HCF of 720 and 405 to find the maximum quantity of milk to be filled in one glass.
405 = 3^{4} × 5
720 = 2^{4} × 3^{2} × 5
HCF = 3^{2} × 5 = 45 ml = Capacity of glass
No. of glasses filled from 1st vessel = = 16
No. of glasses filled from 2nd vessel = = 9
Total number of glasses = 25
જવાબ : To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.
10 = 2 × 5
16 = 2^{4}
20 = 2^{2} × 5
LCM = 2^{4} × 5 = 16 × 5 = 80 minutes
They will start preparing a new card together after 80 minutes.
જવાબ : First, we find HCF of 6339 and 6341 by Euclid’s division method.
6341 > 6339
6341 = 6339 × 1 + 2
6339 = 2 × 3169 + 1
2 = 1 × 2 + 0
HCF of 6341 and 6339 is 1.
Now, we find the HCF of 134791 and 1
134791 = 1 × 134791 + 0
HCF of 134791 and 1 is 1.
Hence, the HCF of the given three numbers is 1.
જવાબ : x = p^{2}q^{3} and y = p^{3}q
LCM = p^{3}q^{3}
HCF = p^{2}q …..(i)
Now, LCM = p^{3}q^{3}
⇒ LCM = pq^{2} (p^{2}q)
⇒ LCM = pq^{2} (HCF)
Yes, LCM is a multiple of HCF.
Explanation:
Let a = 12 = 2^{2} × 3
b = 18 = 2 × 3^{2}
HCF = 2 × 3 = 6 …(ii)
LCM = 2^{2} × 3^{2} = 36
LCM = 6 × 6
LCM = 6 (HCF) …[From (ii)]
Here LCM is 6 times HCF.
જવાબ : Let n, n + 1, n + 2 be three consecutive positive integers.
We know that n is of the form 3q, 3q + 1, or 3q + 2.
Case I. When n = 3q,
In this case, n is divisible by 3,
but n + 1 and n + 2 are not divisible by 3.
Case II. When n = 3q + 1,
In this case n + 2 = (3q + 1) + 2
= 3q + 3
= 3(q + 1 ), (n + 2) is divisible by 3,
but n and n + 1 are not divisible by 3.
Case III.
When n = 3q + 2, in this case,
n + 1 = (3q + 2) + 1
= 3q + 3 = 3 (q + 1 ), (n + 1) is divisible by 3,
but n and n + 2 are not divisible by 3.
Hence, one and only one out of n, n + 1 and n + 2 is divisible by 3.
જવાબ : 306 = 2 × 3^{2} × 17
657 = 32 × 73
HCF = 3^{2} = 9
LCM = 2 × 3^{2} × 17 × 73 = 22338
L.H.S. = LCM × HCF = 22338 × 9 = 201042
R.H.S. = Product of two numbers = 306 × 657 = 201042
L.H.S. = R.H.S.
જવાબ : Let a be a positive odd integer
By Euclid’s Division algorithm:
a = 4q + r …[where q, r are positive integers and 0 ≤ r < 4]
a = 4q
or 4q + 1
or 4q + 2
or 4q + 3
But 4q and 4q + 2 are both even
a is of the form 4q + 1 or 4q + 3.
જવાબ : Let x be any positive integer and y = 3.
By Euclid’s division algorithm; x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3) Therefore, x = 3q, 3q + 1 and 3q + 2 As per the given question, if we take the square on both the sides, we get; x^{2} = (3q)^{2} = 9q^{2} = 3.3q^{2} Let 3q^{2} = n Therefore, x^{2} = 3n ………………….(1) x^{2} = (3q + 1)^{2} = (3q)^{2} + 1^{2} + 2 × 3q × 1 = 9q^{2} + 1 + 6q = 3(3q^{2} + 2q) + 1 Substitute, 3q^{2}+2q = n, to get, x^{2} = 3n + 1 ……………………………. (2) x^{2} = (3q + 2)^{2} = (3q)^{2 }+ 2^{2 }+ 2 × 3q × 2 = 9q^{2} + 4 + 12q = 3(3q^{2} + 4q + 1) + 1 Again, substitute, 3q^{2 }+ 4q + 1 = m, to get, x^{2} = 3n + 1…………………………… (3) Hence, from eq. 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3n or 3n + 1 for some integer n.જવાબ : (i) 1400
140 = 2 × 2 × 2 × 5 x 5 × 7 × 1
1 
Rational No 
A 
Root of 1 
2 
Irrational No. 
B 
Rational & Irrational no. 
3 
Real No. 
C 
1/3 
4 
NonReal No. 
D 
Square root of 2 
જવાબ :
1C , 2D, 3B, 4A
1 
Whole No. 
A 
1/7 
2 
Natural No 
B 
0,1,2,3…… 
3 
Integers. 
C 
1,2,3,4…… 
4 
Fractions 
D 
…2,1,0,1,2… 
જવાબ :
1B, 2C, 3D, 4A
1 
Even No. 
A 
4,6,8,9,12 
2 
Odd No . 
B 
2,3,5,7,11 
3 
Prime No. 
C 
1,3,5,7,11 
4 
Composite No. 
D 
2,4,6,8,10 
જવાબ :
1D, 2C, 3B, 4A

No. 

H.C.F. 
1 
6, 12, 15 
A 
1 
2 
4, 8, 16 
B 
3 
3 
10, 25, 50 
C 
4 
4 
2, 3, 4 
D 
5 
જવાબ :
1B,. 2C, 3D, 4A

No. 

L.C.M 
1 
12, 13 
A 
150 
2 
100, 15 
B 
156 
3 
50, 60 
C 
66 
4 
2, 33 
D 
300 
જવાબ :
1B, 2A, 3D, 4C
1 
Rational No 
A 
√2 
2 
Irrational No. 
B 
√2 & √4 
3 
Real No. 
C 
√4 
4 
NonReal No. 
D 
√4 
જવાબ :
1D, 2A, 3B, 4C
1 
Even No. 
A 
32,34,35,36 
2 
Odd No . 
B 
31,37,41,43 
3 
Prime No. 
C 
32,34,36 
4 
Composite No. 
D 
31,33,35,37 
જવાબ :
1C, 2D, 3B, 4A

No. 

H.C.F. 
1 
33,100 
A 
4 
2 
44, 46 
B 
3 
3 
3, 39 
C 
2 
4 
8,12 
D 
1 
જવાબ :
1D, 2C, 3B, 4A

No. 

L.C.M. 
1 
33,100 
A 
24 
2 
44, 46 
B 
3300 
3 
3, 39 
C 
903 
4 
8,12 
D 
39 
જવાબ :
1B, 2C, 3D, 4A

No. 

H.C.F. 
1 
12, 13 
A 
5 
2 
100, 15 
B 
2 
3 
50, 60 
C 
1 
4 
2, 32 
D 
10 
જવાબ :
1C,2A,3D,4B
Chapter 01 : Real Numbers
.આ પ્રકરણને લગતા વિવિધ એનિમેશન વિડીયો, હેતુલક્ષી પ્રશ્નો, ટૂંકા પ્રશ્નો, લાંબા પ્રશ્નો, પરિક્ષામાં પુછાઈ ગયેલા પ્રશ્નો તેમજ પરિક્ષામાં પુછાઈ શકે તેવા અનેક મુદ્દાસર પ્રશ્નો જોવા અમારી વેબસાઈટ પર રજીસ્ટર થાઓ અથવા અમારી App ફ્રી માં ડાઉનલોડ કરો.
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.
For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.